area of isosceles triangle?

Question

How do I calculate the area of an isosceles triangle?
I have:
-The measurement of the one side that is unequal.
-The angle on the opposite corner (opposite of the unequal side).

soccer stopper; the length is (p-q) and the angle is theta. (i.e., variables)

thin, i *don't have the height.
i have:
-length of one side: (p-q)
-angle of opposite sise: (variable: theta)
I do not have the height. It would be best as a function of theta.

Answer 1

You know the vertex angle is x degrees
So each base angle must be (180-x)/2 degrees.
You know the base is = b.
Draw an altitude (h) from the vertex to the base. It will be perpendicular to the base and also bisect the base.

b/(2h) = tanx/2
So h = b/(2tanx/2)
So A = b^2/(4tanx/2)

Answer 2

Using the angle you know, you can calculate the height of the triangle...

Do this "thought experiment": stand the triangle on its 'unequal' side. Drop a line from the apex at the top to the middle of the base. Now you have two "mirror-image" right triangles. Pick one up and swing it all the way around so that its hypotenuse is touching the hypontenuse of the other triangle. Now you have a simple rectangle, with one side equal to the height of your original isoseles triangle, and one side equal to half the length of the base of your original isoseles triangle. You know how to calculate the area of a rectangle, right?

Answer 3

x= known angle
y= the length of the unequal side
z= the perpendicular height
Split the triangle into 2 right angle triangles.
Find z:
(y/2)/ Tan(x/2)
Find the area:
zy/2

Answer 4

since you have the angle u can calculate the other 2 angles ( they are equal ) , draw perpendicular from the given corner so you ll get a right angle triangle where you are given base ( half of the side that is unequal), so calculate the equals sides using sin , n the are woudl be 0.5 * (equal side squared) * sin (the angle between the equal sides).

Answer 5

It would be helpful if you told me what the angles and measurements are.